Question

y=y(x)是由sin(xy)=ln[(x+e)/y]+1确定的隐函数，则y'(0)=?

Answer 1

简单计算一下即可，答案如图所示

Answer 2

因为
e的x+y次+sin(xy)=1，两边求导，e的x+y次（x′+y′）+cos（xy）（x′y+xy′）=0
由于e的x+y次与cos（xy）不可能为零。则x′+y′=0，x′y+xy′=0
解出y′=y²/x，那么
y(x)的导数是就为y²/x
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Answer 3

sin(xy)=ln[(x+e)/y]+1
两边对x求导，cos(xy)*(y+x*y’)=y/(x+e)*[y-(x+e)*y’]/y^2,
cos(xy)*y+
cos(xy)*x*y’=1/(x+e)*[y-(x+e)*y’]/y,
cos(xy)*y+
cos(xy)*x*y’=1/(x+e)-y’/y,
cos(xy)*x*y’+y’/y
=1/(x+e)
-cos(xy)*y,
(cos(xy)*x*y+1)y’
=y/(x+e)
-cos(xy)*y^2
y’
=[y/(x+e)
-cos(xy)*y^2]/
(cos(xy)*x*y+1)
把x=0代入原方程得：ln[(0+e)/y]+1=1-lny+1=2-lny=0,lny=2,y=e^2
把x=0,y=e^2代入导数式中得：
y’(0)=
[e^2/(0+e)
-cos(0)*
e^4]/
(cos(0)*0*
e^2+1)
=[e-e^4]/1
=e-e^4